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Chapter 8: Q 4E (page 469)

For the conditions of Exercise 2, how large a random sample must be taken in order that\({\bf{P}}\left( {{\bf{|}}{{{\bf{\bar X}}}_{\bf{n}}}{\bf{ - \theta |}} \le {\bf{0}}{\bf{.1}}} \right) \ge {\bf{0}}{\bf{.95}}\)for every possible value of胃?

Short Answer

Expert verified

The needed sample size is \( \ge 1537\)

Step by step solution

01

Given information

The given information is \(P\left( {|{{\bar X}_n} - \theta | \le 0.1} \right) \ge 0.95\)

02

Finding sample size

Here \({\bar X_n}\) has the normal distribution with mean \(\theta \) and variance \(4/n\) . Hence, the random variable \(Z = \left( {{{\bar X}_n} - \theta } \right)/\left( {2/\sqrt n } \right)\) will has standard normal distribution.

Therefore,

\(\begin{align}\Pr \left( {|{{\bar X}_n} - \theta | \le 0.1} \right) &= \Pr \left( {|Z| \le 0.05\sqrt n } \right)\\ &= 2\Phi \left( {0.05\sqrt n } \right) - 1\end{align}\)

Here given that,

\(\begin{align}\Pr \left( {|{{\bar X}_n} - \theta | \le 0.1} \right) \ge 0.95\\2\Phi \left( {0.05\sqrt n } \right) - 1 \ge 0.95\\\Phi \left( {0.05\sqrt n } \right) \ge 1.95/2\\0.05\sqrt n \ge {\Phi ^{ - 1}}\left( {0.975} \right)\\0.05\sqrt n \ge 1.96\,(It\,found\,from\,table\,value\,of\,\Phi )\\n \ge \frac{{{{1.96}^2}}}{{{{0.05}^2}}}\\n \ge 1536.64\end{align}\)

The required sample size is\( \ge 1537\)

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Most popular questions from this chapter

Suppose that X1,鈥︹赌�.,Xnform a random sample from the exponential distribution with unknown parameter 尾. Construct an efficient estimator that is not identically equal to a constant, and determine the expectation and the variance of this estimator.

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Suppose that \({{\bf{X}}_{\bf{1}}}{\bf{, }}{\bf{. }}{\bf{. }}{\bf{. , }}{{\bf{X}}_{\bf{n}}}\) form a random sample from the normal distribution with unknown mean \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\), and also that the joint prior distribution of \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\) is the normal-gamma distribution satisfying the following conditions: \({\bf{E}}\left( {\bf{\tau }} \right){\bf{ = 1,Var}}\left( {\bf{\tau }} \right){\bf{ = }}\frac{{\bf{1}}}{{\bf{3}}}\,\,\,{\bf{and}}\,\,{\bf{Pr}}\left( {{\bf{\mu > 3}}} \right){\bf{ = 0}}{\bf{.5}}\,\,{\bf{and}}\,\,{\bf{Pr}}\left( {{\bf{\mu > 0}}{\bf{.12}}} \right){\bf{ = 0}}{\bf{.9}}\,\)

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Determine whether or not each of the five following matrices is orthogonal:

  1. \(\left( {\begin{align}{\bf{0}}&{\bf{1}}&{\bf{0}}\\{\bf{0}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{0}}\end{align}} \right)\)
  2. \(\left( {\begin{align}{{\bf{0}}{\bf{.8}}}&{\bf{0}}&{{\bf{0}}{\bf{.6}}}\\{{\bf{ - 0}}{\bf{.6}}}&{\bf{0}}&{{\bf{0}}{\bf{.8}}}\\{\bf{0}}&{{\bf{ - 1}}}&{\bf{0}}\end{align}} \right)\)
  3. \(\left( {\begin{align}{{\bf{0}}{\bf{.8}}}&{\bf{0}}&{{\bf{0}}{\bf{.6}}}\\{{\bf{ - 0}}{\bf{.6}}}&{\bf{0}}&{{\bf{0}}{\bf{.8}}}\\{\bf{0}}&{{\bf{0}}{\bf{.5}}}&{\bf{0}}\end{align}} \right)\)
  4. \(\left( {\begin{align}{}{{\bf{ - }}\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}\\{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{{\bf{ - }}\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}\\{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}&{{\bf{ - }}\frac{{\bf{1}}}{{\sqrt {\bf{3}} }}}\end{align}} \right)\)
  5. \(\left( {\begin{align}{}{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}\\{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}\\{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}\\{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{\frac{{\bf{1}}}{{\bf{2}}}}&{{\bf{ - }}\frac{{\bf{1}}}{{\bf{2}}}}\end{align}} \right)\)

Show that two random variables \({\bf{\mu }}\,\,{\bf{and}}\,\,{\bf{\tau }}\)cannot have the joint normal-gamma distribution such that

\({\bf{E}}\left( {\bf{\mu }} \right){\bf{ = 0}}\,\,{\bf{,Var}}\left( {\bf{\mu }} \right){\bf{ = 1,E}}\left( {\bf{\tau }} \right){\bf{ = }}\frac{{\bf{1}}}{{\bf{2}}}\,\,{\bf{and}}\,\,{\bf{Var}}\left( {\bf{\tau }} \right){\bf{ = }}\frac{{\bf{1}}}{{\bf{4}}}\)
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